Problem: $\begin{cases}a(1)=-11\\\\ a(n)=a(n-1)\cdot 10 \end{cases}$ What is the $4^{\text{th}}$ term in the sequence?
Solution: This is a recursive formula. It tells us that the first term is $-11$ and that the common ratio is $10$. $\begin{aligned} {a(1)}&=-11 \\\\ {a(2)}&={a(1)}\cdot 10=-110 \\\\ {a(3)}&={a(2)}\cdot 10=-1100 \\\\ {a(4)}&={a(3)}\cdot 10=-11{,}000 \end{aligned}$ The $4^{\text{th}}$ term is $-11{,}000$.